Does anyone know if this function has a name? I came up with it by looking at the power series for $e^z$, changing the summation to an integral, and substituting the gamma function for the factorial function.

@mlbaker I recalled this equation from one of my bookmarks: artofproblemsolving.com/Forum/viewtopic.php?p=1226810#p1226810 . I also found another source which you may find useful: Erdelyi, A., et al., Higher Transcendental Functions, vol 3. You can find it online at apps.nrbook.com/bateman/Vol3.pdf To be more specific look at the page 217 where author introduces your integral and denotes it by $\nu (x)$. It may be a good place to start looking for other references I guess.
–
qoqoszJun 17 '12 at 19:21

1

@mlbaker: This may be a "troll function," as you call it, but it is pretty interesting nonetheless!
–
user26872Jun 18 '12 at 18:46

We consider the function for $x\ge 0$.
As found by @Mathlover, it is not hard to get the Laplace transform of $f(x)$.
Inverting, we find
$$f(x) = \frac{1}{2\pi i} \int_\gamma ds\, e^{s x}\frac{1}{s\log s},$$
where $\gamma$ is the Bromwich contour.
There is a singularity at $s=1$ and a cut from $s=0\to -\infty$.
We deform the contour in the usual way and simplify, with the result
$$\begin{eqnarray*}
f(x) &=& e^x - \int_0^\infty d\rho\,
\frac{e^{-\rho x}}{\rho(\log^2\rho + \pi^2)} \\
&=& e^x - \int_{-\infty}^\infty d\sigma\,
\frac{e^{-x e^\sigma}}{\sigma^2 + \pi^2}.
\end{eqnarray*}$$
It is possible to go further with the last integral, but we'll stop here.

Clearly the derivative of $\Delta(x)$ doesn't exist at $x=0$
(the integral
$\int_0^\infty d\sigma\,\frac{\cosh \sigma}{\sigma^2+\pi^2}$ diverges)
so there is no Maclaurin series for $f(x)$.
We can do a Taylor series about any $x>0$.
Of course, we must evaluate integrals of the form
$$\left.\frac{d^k}{d x^k} \Delta(x)\right|_{x=x_0} = (-1)^k \int_0^\infty d\sigma\,
\frac{e^{k\sigma -x_0 e^\sigma} + e^{-k\sigma -x_0 e^{-\sigma}}}{\sigma^2 + \pi^2}.$$
These integrals are uniformly convergent for $x_0 >0$.
In this regard it is useful to notice that
$e^{-k\sigma -x_0 e^{-\sigma}} \le 1$ and
$$e^{k\sigma -x_0 e^\sigma} \le
\begin{cases}
e^{-x_0}, & x_0 \ge k \\
\left(\frac{k}{x_0 e}\right)^k, & \mathrm{else}.
\end{cases}$$
For large $x_0$ the dominant terms in the expansion come from the series for $e^x$, as expected.

Trapezoidal rule

Approximate the integral by a sum.
Let the step size be $1/m$, where $m=1,2,\ldots$, and let
$g_k = \frac{x^{k/m}}{\Gamma(k/m+1)}$.
Then,
$$\begin{eqnarray*}
\int_0^\infty dt\, \frac{x^t}{\Gamma(t+1)} &\simeq&
\lim_{n\to\infty}
\left(
\frac{1}{m}\sum_{k=0}^n g_k -\frac{1}{2m}(g_0+g_n) \right)\\
&=& \frac{1}{m}E_{1/m}(x^{1/m}) -\frac{1}{2m} \\
&=& \frac{1}{m} e^x\left(1+\sum_{k=1}^{m-1}\frac{\gamma(1-k/m,x)}{\Gamma(1-k/m)}\right) - \frac{1}{2m},
\end{eqnarray*}$$
where $E_\alpha(z)$ is the Mittag-Leffler function and $\gamma$ is the lower incomplete gamma function.
(The relation for $E_{1/m}(x^{1/m})$ used above can be found in this paper.)

In the limit $m\to\infty$ the sum is equal to the original integral, so
$$\begin{eqnarray*}
f(x) &=& e^x \lim_{m\to\infty} \frac{1}{m} \sum_{k=1}^{m-1}\frac{\gamma(1-k/m,x)}{\Gamma(1-k/m)} \\
&=& e^x \int_0^1 ds\,\frac{\gamma(s,x)}{\Gamma(s)}
\end{eqnarray*}$$
This is the integral formula found by @sos440 using another method.

I don't think the function has any specific name. One plausible reason why this function doesn't have its own name is probably because it's growth is very closely related to $\exp(x)$ as expected. A very rough analysis as shown below gives some growth of this function with respect to $x$. The argument can be improved to get better growth rates of the function.